If-1-a-2-1-b-2-1-c-2-1-ab-1-bc-1-ca-then-prove-that-a-b-c-

Question Number 70312 by Shamim last updated on 03/Oct/19

If, \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} then prove

that, a = b = c .

Commented by MJS last updated on 03/Oct/19

this is true for a, b, c \in R

{it}^{'} s false for a, b, c \in C

b = a (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) \land c = a (- \frac{1}{2} + \frac{\sqrt{3}}{2} i)

(\begin{matrix} a \\ b \\ c \end{matrix}) = a \times (\begin{matrix} 1 \\ - \frac{1}{2} + \frac{\sqrt{3}}{2} i \\ - \frac{1}{2} - \frac{\sqrt{3}}{2} i \end{matrix}) is a solution

(\begin{matrix} a \\ b \\ c \end{matrix}) = (\begin{matrix} a \\ b \\ \frac{a b (a + b) \pm \sqrt{3} a b (a - b) i}{2 (a^{2} - a b + b^{2})} \end{matrix}) is a solution

Answered by $@ty@m123 last updated on 03/Oct/19

L e t \frac{1}{a} = x, \frac{1}{b} = y, \frac{1}{c} = z

\Rightarrow x^{2} + y^{2} + z^{2} = x y + y z + z x

\Rightarrow x^{2} + y^{2} + z^{2} - (x y + y z + z x) = 0

\Rightarrow 2 {x^{2} + y^{2} + z^{2} - (x y + y z + z x)} = 0

\Rightarrow {(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2} = 0

\Rightarrow x - y = 0, y - z = 0, z - x = 0

\Rightarrow x = y = z

\Rightarrow a = b = c